Math, asked by rajuyadav56, 10 months ago

prove that sin0-cos0+1/sin0+cos0-1=1/sec0-tan0

Answers

Answered by sandy1816
7

Step-by-step explanation:

sinO-cosO+1/sinO+cosO-1

devide by cosO in each term

tanO+secO-1/tanO-secO+1

={(tanO+secO)-1}(tanO-secO)/(tanO-secO+1)(tanO-secO)

=tan²O-sec²O-tanO+secO/(tanO-secO+1)(tanO-secO)

=-1-tanO+secO/(tan-secO+1)(tanO-secO)

=-1/tanO-secO

=1/secO-tanO

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